Different way to solve $|x-3|<|2x|$

Question

I know two ways I can solve $|x-3|<|2x|$

1. By squaring both sides
2. By interpreting the inequality as a statement about distances on the real line.

Question: How can I solve this inequality algebraically, but without squaring?

Thanks!

Answer 1

By a discussion based on the sign rule for the absolute value:

$$t\ge0\implies|t|=t,t\le0\implies |t|=-t.$$

Here you have three cases

$$\begin{align}x\le0&\implies |x-3|<|2x|\equiv -x+3<-2x\\ 0\le x\le3&\implies |x-3|<|2x|\equiv -x+3<2x\\ x \ge3&\implies |x-3|<|2x|\equiv\ \ \ x-3<2x.\end{align}$$

Then you solve the three inequations and match the solutions against the three sub-domains.

Answer 2

Check for critical points i.e. $x-3=0$ and $2x=0$ this will give you $x=3$ and $x=0$, thus your real lines is divided into 3 parts,$(-\infty,0)$,$[0,3)$,$[3,+\infty)$, now solve your equation in each of these intervals.

Answer 3

You should consider the following cases

$$\left\{ \matrix{ x \ge 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,x - 3 < 2x\, \hfill \cr 0 < x < 3\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\, - (x - 3) < 2x \hfill \cr x \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\, - (x - 3) < - 2x \hfill \cr} \right.$$